Firstly, the magic square and 5-sums in box 5 immediate give us the 258 diagonal, some 13 pairs inside the box and 24 pairs outside.
The X-sums on the 24 pairs give us the 6 in box 6 and 8 in box 8.
The magic square in box 5 fully resolves. Easy, right?

Top tip - it's worth colouring known and potential 2s, 4s, 6s, and 8s, as you will want to revisit them many times.

The 5-sums in column 9 can be marked up as a [1234] pair and a [12]/[34] pair.
The 5-sum in column 1 is also a [12]/[34] pair.
The 5-sum in row 1 is a [24]/[13] pair, as 2 is excluded from top-left in a box.
If the 4 in column 9 is in row 7, the [1234] quad in box 3 resolves to a [23] pair, and the 4 in row 1 is pushed into column 6. This puts 2 4s top right of a box, so impossible.
=> The [1234] quad in box 3 is the [14] pair.
=> 2 is above 3 in column 9's 5-sum
=> 2 and 3 can be placed in row 1's 5-sum
=> The 5-sum in column 1 can only be 1 above 4
=> The X-sum in column 1 is 4 above 6.
=> The X-sum in row 5 can only 3x7.
=> row 5 in box 4 must have 6 chaperone the 2 and 8.
=> The 4 in box 4 is in r4c3
=> The 4 in box 1 is in r1c2
=> The 6 in box 1 is in r2c3
=> The 6 in box 3 is in r1c8
Box 6 can be pencil-marked, noting that as the highly-restricted 1 cannot form a domino with a 9.

The 4 in box 8 can only be in row 9
=> The 4 in box 9 can only be in r8c8
=> The 6 in box 9 can only be in r9c9 by absense of X-sum.
=> By the lack of 5-sum the X-sum in row 8 must be 3-7

Column 5 can be pencil-marked:
As r9c5 can't be 6, r3c5 can't be 4, and they must sum to 10 
r2c5 can't be 4, because it would abut a 6
=> r2c5 can't be a 6, as it would abut a 4.
=> the X-sum in box 2 is a [19] pair
=> r3c5 and r9c5 become 6 and 4
=> r2c6 is the 4
=> the 5-sum in box 3 resolves to 1 above 4
=> The X-sum in box 2 resolves to 1 above 9.

To avoid X-sum, column 6 can only have its 7 in r3c7, leaving a [59] pair in box8.
This fully resolves box 2, and most of box 8.
This fully resolves all of column 9.

Given that the 2 in box 9 must be in row 9, the 8 must be in row 7
=> The 8 in box 2 forces box 9's 8 into r7c8
=> The 8 in box 7 must be in row 9
The [15] pairs in column 7 places the 2 in r9c7

The 15 pair in box 6 places the 9 in r6c8
=> The 9 in box 3 is in r3c7
=> The 8 is in r2c7
=> The 2 is in r3c8 to avoid an X-sum
=> 7 in r2c8 completes box 3

Boxes 1 and 7 x-wing the 8 out of column 1 in box 4, and into r5c3
=> r5c1 is 2
=> 2 and 8 are placed in box 1
=> 2, 5, and 8 are all placed in box 7

Everything else just falls instantly at that point - if you did things my way, the 3 in the corner appears quite late on.